Lecture Notes in Differential Equations - Bruce E. Shapiro

67
Now compare equation (9.2) with (9.17). The earlier equation says that
M(t, y)dt + N(t, y)dy = 0 (9.18)
This means that if there is some function φ such that
M = ∂φ
∂t
and N =
∂φ
∂y
Then φ = C is a solution of the differential equation.
(9.19)
So how do we know when equation (9.19) holds? The answer comes from
taking the cross derivatives:
∂M
∂y = ∂2 φ
∂y∂t = ∂2 φ
∂t∂y = ∂N
∂t
(9.20)
The second equality follows because the order of partial differentiation can
be reversed.
Theorem 9.1. If
then the differential equation
∂M
∂y = ∂N
∂t
(9.21)
M(t, y)dt + N(t, y)dy = 0 (9.22)
is called an exact differential equation and the solution is given by some
function
φ(t, y) = C (9.23)
where
M = ∂φ
∂t
and N =
∂φ
∂y
We illustrate the method of solution with the following example.
Example 9.2. Solve
This is in the form Mdt + Ndy = 0 where
(9.24)
(2t + y 2 )dt + 2tydy = 0 (9.25)
M(t, y) = 2t + y 2 and N(t, y) = 2ty (9.26)
First we check to make sure the equation is exact:
∂M
∂y = ∂ ∂y (2t + y2 ) = 2y (9.27)